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ivann1987 [24]
3 years ago
13

Write an equation for the amount of money, , that will be collected if boxes of chocolate bars are sold. Which is the independen

t variable and which is the dependent variable
Mathematics
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

Kindly check explanation

Step-by-step explanation:

Let amount of money = C

If the number of chocolate bars sold = x

Let cost per chocolate bar = p, since cost per bar isn't stated

Amount of money that will be collected :

C(x) = P * x

The dependent variable is the Amount of money that will be collected as the amount of money depends on the number of chockate bars sold

The number of chocolate bars sold us the independent variable as it dictates the amout of money to be collected, hence it is the predictor variable.

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Slope form: y=mx + b
slope is 2/5, fill into equation
Y = 2/5x + b
Plug in point to find b
1 = 2/5(0) + b
1 = 0 + b, b = 1
The answer is A. Y = 2/5x + 1
8 0
3 years ago
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masha68 [24]

Answer:

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Step-by-step explanation:

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krek1111 [17]

Answer:

83.3%

Step-by-step explanation:

The probability of rolling a 3 on a 6-sided dice is 1/6

Since you don't want a three, there's a 5/6 chance of not rolling a 3

5/6 (use calculator and divede 5 by 6) = 0.833333 (move decimal 2 to the right) = 83.3%

Therefore, the most reasonable answer would be C, 83.3%

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3 years ago
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Rudiy27

Answer:

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Step-by-step explanation:

6 0
3 years ago
I need help finding the area
aniked [119]
<h3>Given :</h3>
  • Base of triangle = 7 yd
  • Height of triangle = 10 yd

\\  \\

<h3>To find:</h3>
  • Area of triangle

\\  \\

We know:-

When base and height of triangle is given we use this formula:

\bigstar \boxed{ \rm Area \: of \: triangle =  \frac{base \times height}{2} }

\\  \\

So:-

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{base \times height}{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 10}{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times2 }{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times\cancel2 }{\cancel2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times1 }{1}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5

\\  \\

\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2}  \\

\\  \\

\therefore  \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} {  \red{\bf{yd} }^{ \red2} }}

\\  \\

<h3>know more :-</h3>

\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}]

7 0
2 years ago
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